A chemistry topic in the first year of high school must be written about the process

The amount of gaseous substances mixed n=v vm=

The amount of cu species participating in the reaction n = m m = that....Are you sure that's right? Since the copper is completely dissolved, i.e., all of it is oxidized to Cu2+, Cu loses electrons in total, and all of these electrons enter the NO2 and NO2.

Let the amount of no substance x in the gas mixture, and the amount of no2 substance y be x+y=

Again: n decreases by 3 valence from Hno3 to NO, and 3mol electrons are obtained per mol n; From Hno3 to No2, the valence is reduced by 1, and 1mol of electrons is obtained per mol of N, and these electrons are provided by Cu.

Hence 3x+y=

The solution is x=, y=

Therefore, the NO2 volume v=NVM=, NO2 volume v=NVM=, and the question is less conditional ......At the very least, you have to tell the volume of the NaOH solution.

The equations for copper and concentrated dilute hydrochloric acid are written, and the quantities of substances reflected in the dilute HCl are m and n respectively, and the simultaneous equation is solved.

The second question is the same.

Fe + 6Hno3 = Fe (NO3) 3 + 3 NO2 + 3H2O3Fe + 8H No3 = 3Fe (NO3) 2 + 2 No + 4H2O corresponds, 1792ml corresponds to the corresponding mixture of No and No2.

Since Fe has surplus, it is Fe2+ in the solution, so the N element loses electrons in total.

And there are also NO3-nitrate ions in the solution.

Let the quantity of the substance of no be x, then the quantity of the substance of no2 is.

3x+, the solution is x=

So no for, no 2 for.

Bringing these two numbers into the first two equations, it is calculated that the total amount of Hno3 is and the nitrate left in the solution is, so the ratio of the mass of the reduced nitric acid to the unreduced nitric acid is::5, the amount of the substance concentration of the original nitric acid solution = , the composition of the mixed gas is no and no2, and the volume ratio is no:no2=:

1It's all my hand, you have to!

The final solute generated is ferrous nitrate Fe(NO3)2, in which the mass of iron is, then nitrate is present.

There are の and の2 in total.

According to the conservation of n, nitric acid is shared.

The concentration of orthonitric acid is:

5fe + 14hno3 = 5fe（no3）2 + no2 + 3no + 7h2o

The ratio of reduced nitric acid to unreduced nitric acid is (1+3) (5*2)=2:5, and the mixed gas is NO2 and NO, and the volume ratio is 1:3

H is +1 valence, so b is -3 valence, and the most ** of b can be obtained according to the outermost 8-pip saturation state is +5 valence, so the chemical formula of the oxide can be obtained as B2O5, the relative atomic mass of the O element is 16, and the relative atomic mass of b is X

Calculation: 2x (2x+5*16) =

x = 31

The relative mass of protons and neutrons in the element can be regarded as 1, and the electrons are negligible, so the number of protons can be calculated as y y+y+1 = 31

y = 15

So the element with a relative atomic mass of 31 and a proton number of 15 is boron, and the element symbol p

The chemical formula of the gaseous hydride of a non-metallic element B is BH3, which is known to be a group III element, the most **** oxide is +5 valence, and the most ** oxide of B is B2O5

column, mb 2 (mb 2 + 16 5) =

MB=31 Since the number of protons in the nucleus of b is 1 less than the number of neutrons, and the relative atomic mass of b is 31, the number of protons is 15, which is phosphorus.

The white precipitate is a mixture of baso4 and baco3 plus nitric acid, and the precipitate is baso4

So there is m(baco3)=

n(baco3)=

n(baso4)=

1) S conservation:

na2so4---s---baso4

Therefore n(Na2SO4)=n(BaSO4)=C(Na2SO4)=

2) C conservation.

baco3---c---co2

n(co2）=n(baco3)=v=